A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
思路:动态规划,刚开始想用递归解决结果超时,计算组合数也没有很好的方法。然后参考他人想法用动态规划,但是我的想法是用一个m*n数组存储,空间复杂度太高,下列代码为用一个大小为n的数组存储中间结果,时间复杂度为o(m*n)。
代码:
int uniquePaths(int m, int n)
{
if(m<=0 || n<=0)
{
return 0;
}
if(m==1 || n==1)
{
return 1;
}
int *map=new int[n];
for(int i=0; i<n; ++i)
{
map[i]=1;
}
for(int i=1; i<m; ++i)
{
for(int j=n-2; j>=0; --j)
{
map[j]+=map[j+1];
}
}
return map[0];
}